The general displacement of a simple harmonic oscillator is . Let be its time period. The slope of its potential energy (U) – time (t) curve will be maximum when The value of is ________. [2023]

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Solution

Q.
The general displacement of a simple harmonic oscillator is $x = A \sin ω t$ . Let $T$ be its time period. The slope of its potential energy (U) – time (t) curve will be maximum when $t = \frac{T}{β} .$ The value of $β$ is ________. [2023]

Ans.
(8)

$x = A \sin (ω t)$

$U_{(x)} = \frac{1}{2} k x^{2}$

$\frac{d U}{d t} = \frac{1}{2} k 2 x \frac{d x}{d t}$

$= k A^{2} ω \sin (ω t) \cos (ω t) \times \frac{2}{2}$

${(\frac{d U}{d t})}_{\max} = \frac{k A^{2} ω}{2} {(\sin 2 ω t)}_{\max}$

$2 ω t = \frac{π}{2} \Rightarrow t = \frac{π}{4} ω = \frac{T}{8} \Rightarrow β = 8$

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